1. Problem Set 2 Solutions Tree Reconstruction Algorithms
Marc A. Schaub
February 22nd, 2008
CS 262 Problem Session
Problem Set 2 Solutions
Tree Reconstruction Algorithms
Based on slides by
- Andreas Sundquist and George Asimenos (problem 1)
- Serafim Batzoglou (tree reconstruction)

2. Problem 1(a)

3. Problem 1(b)
Baum-Welch:
Suppose
Forward:
Similar for Backward

4. Problem 1(b)
Baum-Welch:

5. Problem 1(b)
Baum-Welch:

6. Problem 1(b)
Baum-Welch:
Given
Inductive step:
 After training:

7. Problem 1(b)
Viterbi:
Viterbi parse may arbitrarily choose state k over state k’  Akl  Ak’l  a’kl  a’k’l

8. Problem 1(c) akl l=1 2 k=0 1 1/2 Akl l=1 2 k=0 1 ek(b) b=x y k=1 1 2
1/2
Akl
l=1 2
k=0 1
ek(b)
b=x y
k=1 1 2
Ek(b)
b=x y
k=1 3 2 1

9. Problem 1(c) Viterbi akl l=1 2 k=0 1 1/2 x y 1 .9 .045 .3645 .1640 2
1/2 x y 1 .9
.045
.3645
.1640 2
.405
ek(b)
b=x y
k=1 1 2

10. Problem 1(c)
Viterbi x y 1
.75
.1688
.1139
.0769 2
.0375
.0084
.0057
akl
l=1 2
k=0 1
0.9
0.1
ek(b)
b=x y
k=1
0.75
0.25 2
0.5
akl
l=1 2
k=0 1
ek(b)
b=x y
k=1
0.75
0.25 2 ?

11. Additive Distances 1
d1,4 12 4 8 3 7 9 5 11 10 6 2
Given a tree, a distance measure is additive if the distance between any pair of leaves is the sum of lengths of edges connecting them
Given a tree T & additive distances dij, can uniquely reconstruct edge lengths:
Find two neighboring leaves i, j, with common parent k
Place parent node k at distance dkm = ½ (dim + djm – dij) from any node m  i, j

12. Neighbor-Joining Dij = (N – 2) dij – ki dik – kj djk
Guaranteed to produce the correct tree if distance is additive
May produce a good tree even when distance is not additive
Step 1: Finding neighboring leaves
Define
Dij = (N – 2) dij – ki dik – kj djk
Claim: The above “magic trick” ensures that Dij is minimal iff i, j are neighbors 1 3
0.1
0.1
0.1
0.4
0.4 2 4

13. Algorithm: Neighbor-joining
Initialization:
Define T to be the set of leaf nodes, one per sequence
Let L = T
Iteration:
Pick i, j s.t. Dij is minimal
Define a new node k, and set dkm = ½ (dim + djm – dij) for all m  L
Add k to T, with edges of lengths dik = ½ (dij + ri – rj), djk = dij – dik
where ri = (N – 2)-1 ki dik
Remove i, j from L;
Add k to L
Termination:
When L consists of two nodes, i, j, and the edge between them of length dij

14. Parsimony – direct method not using distances
One of the most popular methods:
GIVEN multiple alignment
FIND tree & history of substitutions explaining alignment
Idea:
Find the tree that explains the observed sequences with a minimal number of substitutions
Two computational subproblems:
Find the parsimony cost of a given tree (easy)
Search through all tree topologies (hard)

15. Example: Parsimony cost of one column
Final cost C = 1
{A}
{A, B}
Cost
C+=1 A B A B A A
{A}
{B}
{A}
{A}

16. Parsimony Scoring Given a tree, and an alignment column u
Label internal nodes to minimize the number of required substitutions
Initialization:
Set cost C = 0; node k = 2N – 1 (last leaf)
Iteration:
If k is a leaf, set Rk = { xk[u] } // Rk is simply the character of kth species
If k is not a leaf,
Let i, j be the daughter nodes;
Set Rk = Ri  Rj if intersection is nonempty
Set Rk = Ri  Rj, and C += 1, if intersection is empty
Termination:
Minimal cost of tree for column u, = C

17. Example {B} {A,B} {A} {B} {A} {A,B} {A} A A A A B B A B {A} {A} {A}